Question

The cost of 6 hard-discs equals the cost of 9 printers, the cost of 27 printers equals the cost of 30 scanners, and the cost of 300 scanners equals the cost of 9 computers. If the cost of 3 computers is Rs. 72,000, find the cost of a hard-disc.

• A. Rs. 1800
• B. Rs. 800
• C. Rs. 1500
• D. Rs. 1200

Hint:

Convert all three given relations into one chain connecting hard-discs, printers, scanners and computers through a common multiple.

Answer 1

The Correct Option is D

Step 1: Express each price in terms of the next one.
From $6H = 9P$: $H = \frac{9}{6}P = \frac{3}{2}P$.

Step 2: Bring in the printer-scanner relation.
From $27P = 30S$: $S = \frac{27}{30}P = \frac{9}{10}P$.

Step 3: Bring in the scanner-computer relation.
From $300S = 9C$: $C = \frac{300}{9}S$. Substitute $S = \frac{9}{10}P$:
$$C = \frac{300}{9} \times \frac{9}{10}P = 30P$$
So $P = \frac{C}{30}$.

Step 4: Substitute back to express H directly in terms of C.
$$H = \frac{3}{2}P = \frac{3}{2} \times \frac{C}{30} = \frac{C}{20}$$

Step 5: Use the given value of 3 computers.
Since $3C = 72{,}000$, $C = 24{,}000$. So:
$$H = \frac{C}{20} = \frac{24{,}000}{20} = 1{,}200$$

Final Answer:
The cost of a hard-disc is Rs. 1,200, reached here by expressing every price directly in terms of the computer's cost instead of finding one combined multiple for all four. \[ \boxed{Rs.\ 1200} \]